在我的Minizinc项目中，我正在尝试生成一组n个集合。 给定一组t个不同的数字生成n个不同的集合，其中 基数是在数组m中给出的。 例如： t = 10;并且n = 4;并且m = [3,2,2,3]; 我想生成一组x = [1..3,4..5,6..7,8..10];
但我从下面的代码得到的是 x = [1..3,4..5，{6,10}，7..9]; （我不想使用解决最小化或其他各种解决方案作为我的 目的只是生成一组中间数组。）
int: n = 4; % number of groups
array[1..n] of int: m = [3, 2, 2, 3]; % size of each group
int: t = sum(i in 1..n)(m[i]); % total members
array[1..n] of var set of 1..t: x; % the array of sets
constraint forall(i in 1..n-1)(x[i] > x[i+1]); % SORT .
constraint forall(i in 1..n)(card(x[i] ) = m[i]); % Size of each set
constraint forall(i in 1..n-1)( x[i] intersect x[i+1] = {}); %
% I can't see a way to keep the digits in order
%constraint array_intersect(x) = {}; % this didn't help
solve satisfy;
output [show(x)];
答案 0 :(得分：1)
您可以不受限制地执行此操作。这是一种方法，虽然有点难看：
int: n = 4; % number of sets
array[1..n] of int: s = [3,2,2,3]; % cardinality of the sets
array[1..n] of set of int: x = [ {k | k in sum([s[j] | j in 1..i-1])+1..sum([s[j] | j in 1..i]) } | i in 1..n];
solve satisfy;
constraint true ; % just used to run the model
output [ "x: \(x)\n"];